THE ROYAL ARTILLERY INSTITUTION. 
177 
Formulae are calculated which express x, y, t in terms of 0, <t> and 
the velocity “ u 0 ” that would be attained at the vertex if K (i.e., 2b) 
were constant at the mean value adopted for the portion PQ. 
Another formula gives % i n terms of and again in terms of %. 
Thus, starting from the beginning of the trajectory, where 0 and % 
are known, we can determine the position of the point where the 
inclination is 0'; then treating this as a fresh starting-point, we can 
determine another point in like manner up to the vertex, where 0 = 0. 
The descending branch is similarly calculated from the vertex as a 
first starting-point. 
The horizontal components only of the velocities enter into these 
calculations, but the true velocity at any point can always be found; 
for 
v<f) = . sec 0. 
The proof of the formulae requires the calculus, so those who have 
not learnt it must take the results for granted—viz., equations (2), (6), 
(7), (8). 
Some steps will be omitted in the following, as the whole is given in 
Arts. 61, 62. 
The following elementary formulae must be borne in mind 
C ^~ = v, and ^ =/ are the velocity and acceleration in direction of tangent, Eqns J<?) 
dP 
dx <fix 
di =u ’ anA d¥ 
dt d dP 
u horizontal components of do. do. 
„ vertical „ „ 
In this case f= — 2 bv s . 
* Tifif, « — ^ — 
-j- . w/y . - dx dy . 
Letp = ± = tan +, ^ = cos *, £ = sm +, 
§ 2 . 
§ 9 , 10 . 
ds I 2 . dp I 3 -j , 9 
d= 1 + J =1+ ^ 
Since p = 
dy 
dy dt dp 
dx dx 3 dt 
dt 
dx dPy dy d%x 
dx 
2 
dt 
■(f) 
(D 
The first step in this, and similar dynamical problems, is to state the § 2 . 
* This is evident with a figure. “ ds,” being an infinitesimal portion 
of arc, may be treated as a straight line coinciding with the tangent. 
5i 2 =£j» + %r,||| = 1 + £i • 
